Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, …) which sum to n.
For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.
Java Solution
This is a dp problem. The key is to find the relation which is dp[i] = min(dp[i], dp[i-square]+1). For example, dp[5]=dp[4]+1=1+1=2.
public int numSquares(int n) { int max = (int) Math.sqrt(n); int[] dp = new int[n+1]; Arrays.fill(dp, Integer.MAX_VALUE); for(int i=1; i<=n; i++){ for(int j=1; j<=max; j++){ if(i==j*j){ dp[i]=1; }else if(i>j*j){ dp[i]=Math.min(dp[i], dp[i-j*j] + 1); } } } return dp[n]; } |
other solution could be
generate all squares up to sqrt(value)
and after that, it is just same solution w/ coins